Applications — sphere, cylinder, infinite plane: Visual walkthrough

1.8.7 · D2 · Physics › Electromagnetism › Applications — sphere, cylinder, infinite plane

Hum kya assume kar sakte hain

Pehle symbol se pehle, yeh hai hamara poora toolkit. Agar tum in se nahi mile, yeh Gauss's Law, Electric Flux aur Symmetry in Physics mein build kiye gaye hain.

Formula Woh ek law jis par sab kuch tika hai

$arrows out of the bag \oint_{S} E \cdot d A = charge in the bag \frac{Q _{e n c}}{ε _{0}}$

$\oint_{S}$ — chota circle matlab hai "ek closed surface $S$ par add up karo" (ek surface jiske ek andar aur ek bahar hai, jaise ek balloon).
$d A$ — us surface ka ek tiny patch, ek arrow ki tarah draw kiya gaya jo bag se seedha bahar point karta hai, jiski length patch ke area ke barabar hai.
$E \cdot d A$ — field actually us patch ko kitna pierce karti hai (from Electric Flux).
$Q_{e n c}$ — sirf woh charge jo surface ke andar trapped hai. Bahar ka charge kuch contribute nahi karta.
$ε_{0}$ — nature ka ek fixed constant, $8.85 \times 1 0^{- 12}$ , jo charge aur flux ke beech "exchange rate" set karta hai.

YEH law kyun aur Coulomb's Law kyun nahi? Coulomb ek point ka field deta hai aur tumhe infinitely many pieces add karne par majboor karta hai. Gauss hume woh saari summation skip karne deta hai — agar hum surface ke baare mein clever hain. Woh cleverness symmetry hai, aur sphere wahan sabse zyada shine karti hai.

Step 1 — Source draw karo aur pucho "arrows kis direction mein point kar sakte hain?"

KYA. Hum radius $R$ ki ek ball rakhte hain jisme total charge $Q$ perfectly evenly uske volume mein spread hai, aur hum puuchte hain: kisi baahri point par, $E$ kis direction mein hai?

KYUN. Hum Gauss's law use nahi kar sakte jab tak hum $E$ ki direction aur pattern nahi jaante. Symmetry yeh free mein answer karti hai — kisi bhi math se pehle.

PICTURE. Figure dekho. Socho tum ek field-arrow ho ball se door ek point par khade ho. Kya tum left jhuk sakte ho? Agar left allowed hota, toh poori picture ko rotate karne par (jo kuch nahi badalta — ball har angle se identical dikhti hai) tumhe ek contradiction milti: arrow ko kahin point karna hoga jo koi bhi rotation prefer nahi kar sakta. Woh aisi sirf ek direction hai — centre se seedha bahar — radially outward.

Figure — Applications — sphere, cylinder, infinite plane

Step 2 — Woh bag chuno jo symmetry se match kare

KYA. Hum charge ke around ek imaginary sphere (Gaussian surface) of radius $r$ wrap karte hain, same centre share karte hue.

KYUN. Humne abhi seekha $E$ radially point karta hai aur har radius par ek strength hai. Ek concentric sphere par, har patch $d A$ bhi radially outward point karta hai — toh $E$ aur $d A$ har jagah parallel hain, aur $∣ E ∣$ us par har jagah same hai. Yahi woh double-jackpot hai jo flux integral ko trivial banata hai.

PICTURE. Red dashed sphere hamara bag hai. Us par har point par, black field arrow aur outward patch arrow ek hi line par lie karte hain. Koi grazing nahi, koi variation nahi.

Step 3 — Flux integral collapse karo

KYA. Hum apne chosen sphere par left side $\oint E \cdot d A$ compute karte hain.

KYUN. Kyunki $E ∥ d A$ har jagah hai, dot product plain multiplication $E d A$ ban jaata hai. Kyunki $E$ surface par constant hai, woh sum se bahar aa jaata hai.

PICTURE. Har arrow apne patch ko head-on pierce karta hai (angle $0$ , toh $cos 0 = 1$ ). Saare patches add karne par sirf sphere ka total area milta hai.

$\oint E \cdot d A = \oint E d A = E total area \oint d A = E (4 π r^{2})$

$E \cdot d A \to E d A$ — dot product ka $cos θ$ ban gaya $cos 0 = 1$ kyunki field aur patch parallel hain.
$E$ $\oint$ ke bahar nikala — legal sirf tabhi jab $E$ har patch par same ho (Step 2).
$\oint d A = 4 π r^{2}$ — radius $r$ ki sphere ka surface area; yeh pure geometry hai, koi physics nahi.

Step 4 — Andar ka charge count karo (outside case, $r > R$ )

KYA. Ball se bade bag ke liye, $Q$ ka sab kuch andar hai: $Q_{e n c} = Q$ .

KYUN. Gauss's law sirf isi ki parwaah karta hai jo hamari chosen surface ke andar trapped hai. Jab $r > R$ , poori ball swallow ho jaati hai.

PICTURE. Red ball poori tarah dashed Gaussian sphere ke andar baithti hai — sab kuch enclosed hai.

$Q_{e n c} = Q (r > R)$

Ab Gauss's law ke dono sides equate karo:

$flux, Step 3 E (4 π r^{2}) = charge/ ε_{0} \frac{Q}{ε _{0}} ⟹ E_{o u t} = \frac{1}{4 π ε _{0}} \frac{Q}{r ^{2}}$

Dono sides ko $4 π r^{2}$ se divide karo → $4 π$ familiar Coulomb constant $\frac{1}{4 π ε _{0}}$ mein regroup ho jaata hai.
Result denominator mein $r^{2}$ par depend karta hai → ek inverse-square law.

Step 5 — Bag ko ball ke andar shrink karo ( $r < R$ )

KYA. Ab Gaussian sphere ko ball se chhota hone do. Sirf radius $r$ ke andar ka charge enclosed hai — charge ki outer shell hamara bag ke bahar hai aur ignore ki jaati hai.

KYUN. Gauss phir se: radius $r$ se pare ka charge is surface ke flux mein kuch contribute nahi karta. Humein $Q_{e n c}$ recompute karna hoga.

PICTURE. Dashed Gaussian sphere red ball ke andar baithti hai. Sirf shaded inner core count hota hai; outer rind excluded hai.

Pehle, charge density $ρ$ — charge per unit volume, constant kyunki ball uniform hai:

$ρ = \frac{Q}{\frac{4}{3} π R ^{3}}$

Numerator $Q$ — total charge.
Denominator $\frac{4}{3} π R^{3}$ — ball ka poora volume. Toh $ρ$ = "charge kitna densely packed" hai.

Radius $r$ ke andar ka charge density times chhote volume ke barabar hai:

$Q_{e n c} = ρ \cdot \frac{4}{3} π r^{3}$

$\frac{4}{3} π r^{3}$ — sirf inner sphere ka volume (radius $r$ , $R$ nahi).

Ise Gauss's law mein same flux $E (4 π r^{2})$ ke saath feed karo:

\;\;\Longrightarrow\;\; \boxed{\,E_{in} = \frac{\rho\,r}{3\varepsilon_0} = \frac{Q\,r}{4\pi\varepsilon_0 R^3}\,}$$ - Left side mein $r^2$ hai; enclosed charge mein $r^3$ hai; divide karne par **$r$ ki ek power upar** rehti hai. - Toh andar, $E$ $r$ ke saath **linearly grow** karta hai — jitna gehre jaao, field *utna kamzor*, bilkul centre par zero ho jaata hai. --- ## Step 6 — Degenerate case: exact centre, aur conductor **KYA.** Do edge cases jinhein formula ko survive karna hai: > (a) $r=0$ (dead centre), aur (b) ek hollow conducting shell jahan saara charge surface par rehta hai. **KYUN.** Ek derivation tabhi trustworthy hai jab woh apni extremes par sensibly behave kare. Inhe skip karo aur ek reader ek wall se takraaega jo humne kabhi nahi dikhayi. **PICTURE.** Centre par, Gaussian sphere ko ek dot tak shrink karo: woh koi charge enclose nahi karta, toh $Q_{enc}\to 0$ aur flux $\to 0$. Conductor ke liye, charge ek thin red skin ki tarah baithta hai; skin ke *andar* koi bhi bag kuch enclose nahi karta. ![[deepdives/dd-physics-1.8.07-d2-s06.png]] **(a) Solid ball ka centre.** $r=0$ ko $E_{in}=\dfrac{Qr}{4\pi\varepsilon_0R^3}$ mein daalo → $E=0$. Yeh linear-growth picture se match karta hai: field middle par vanish hoti hai. Physically, charge har direction mein equally pull karta hai, cancel ho jaata hai. **(b) Conducting shell.** Saara charge outer surface par baithta hai. Kisi bhi $r<R$ ke liye, $Q_{enc}=0$, toh $E_{in}=0$ **andar har jagah**, sirf centre par nahi. Dekho [[Conductors in Electrostatics]]. $$E_{in}^{\text{conductor}} = 0 \quad (r<R), \qquad E_{out} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\quad (r>R)$$ > [!mistake] Dono "insides" ko confuse karna > **Solid uniform ball:** andar ka field $r$ ki tarah badhta hai (zero sirf centre par hai). **Hollow conductor:** andar ka field *har jagah* zero hai. **Fix:** pucho "kya charge volume mein spread hai, ya surface par stuck hai?" — woh decide karta hai kaunsa inside-formula use karna hai. --- ## Step 7 — Saare pieces ko ek graph mein stitch karo **KYA.** Uniform ball ke liye $E(r)$ ko saare radii par assemble karo. **KYUN.** Dono regimes ko joined dekhna prove karta hai ki woh boundary $r=R$ par *agree karte hain* — ek final consistency check. **PICTURE.** Rising red line ($E\propto r$) andar, falling black curve ($E\propto 1/r^2$) bahar. Woh exactly $r=R$ par milte hain: field surface par peak karti hai. $$E(R)_{\text{inside}} = \frac{Q\,R}{4\pi\varepsilon_0 R^3} = \frac{Q}{4\pi\varepsilon_0 R^2} = E(R)_{\text{outside}} \checkmark$$ - $r=R$ dono boxed formulas mein daalo → identical. Graph mein surface par **koi jump nahi** hai; dono branches haath milaate hain. ![[deepdives/dd-physics-1.8.07-d2-s07.png]] > [!example] Shape feel karne ke liye numbers > Lo $Q = 2\times10^{-6}$ C, $R = 0.1$ m, aur $\frac{1}{4\pi\varepsilon_0}=9\times10^9$. > - **Surface par** $r=R=0.1$: $E = \dfrac{(9\times10^9)(2\times10^{-6})}{0.1^2} = 1.8\times10^6$ N/C — **peak**. > - **Bahar**, $r=0.2$: $E = \dfrac{(9\times10^9)(2\times10^{-6})}{0.2^2} = 4.5\times10^5$ N/C — peak ka ek chautha (double distance → quarter field, inverse-square). > - **Andar** (uniform ball), $r=0.05$: $E = \dfrac{(9\times10^9)(2\times10^{-6})(0.05)}{0.1^3} = 9\times10^5$ N/C — peak ka exactly **aadha** (aadha radius → aadha field, linear). *Kyun:* $E_{in}=\frac{Qr}{4\pi\varepsilon_0R^3}$ use kiya. --- ## Ek-picture summary ![[deepdives/dd-physics-1.8.07-d2-s08.png]] Sab kuch compressed: radial-symmetry surface chuno → flux $E\cdot4\pi r^2$ tak collapse ho jaata hai → $Q_{enc}/\varepsilon_0$ ke barabar karo → $E$ padhlo. Enclosed charge andar $r^3$ ki tarah badhta hai ($E\propto r$ deta hai) aur bahar $Q$ par freeze ho jaata hai ($E\propto 1/r^2$ deta hai), toh field surface par ek peak tak chadh jaati hai aur phir fade ho jaati hai. > [!recall]- Feynman: plain words mein poori walkthrough > Ek fuzzy red ball of electric "stuff" socho. Field-arrows us se bahar shoot karna chahte hain, lekin kis direction mein? Ball kisi bhi direction mein ghoomne par same dikhti hai, toh arrows ke paas sideways lean karne ka koi reason nahi — woh seedhe bahar fly karte hain, aur unki strength sirf is par depend kar sakti hai ki tum kitna door ho. > > Ab imagine karo ball ke around ek stretchy bubble slip karo. Gauss ka rule kehta hai ki bubble se poking karne wale arrows ki sankhya us ke andar trapped charge ke barabar hai (ek fixed constant times). Kyunki humne ek *round* bubble chuna, har arrow use head-on pierce karta hai aur counting bas "field strength times bubble area = $E\times 4\pi r^2$" hai. > > Agar bubble ball se bada hai, woh *saara* charge trap karta hai — toh jaise tum bahar jaate ho aur bubble badhta hai, arrows ki fixed pile zyada area par thin spread ho jaati hai, $1/r^2$ ki tarah fade karti hai. Isliye ball, bahar se, bilkul ek single dot ki tarah dikhti hai apne centre par. > > Agar bubble *chhota* hai, ball ke andar buried, woh sirf little core mein charge trap karta hai — aur ek chhota core bahut kam charge hold karta hai. Toh centre ke paas field tiny hai; yeh seedhi line ki tarah badhti hai jaise tum bahar jaate ho, surface par sabse strong hit karti hai. Bilkul middle mein: kuch nahi, har direction mein equally pull hota hai. Aur agar ball ek hollow metal shell hai, toh saara charge skin par chhupta hai, toh *andar kahin bhi* dead calm hai — zero field. > [!mnemonic] Ek line mein sphere > **Bahar: point-charge, $1/r^2$. Solid andar: linear, $\propto r$. Hollow andar: zero. Peak skin par.** --- ## #flashcards/physics Charged sphere ke liye $\vec E$ radial kyun hona chahiye ::: Rotational symmetry — ball har angle se identical dikhti hai, toh koi bhi sideways direction special nahi hai Concentric Gaussian sphere se flux ::: $E\cdot 4\pi r^2$ (field constant aur har patch ke parallel) Charged sphere ke bahar field ($r>R$) ::: $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ Uniform solid ball ke andar field ($r<R$) ::: $E=\dfrac{Qr}{4\pi\varepsilon_0R^3}\propto r$ Hollow conductor ke andar field ::: har jagah $0$ Sphere ka field sabse strong kahan hai ::: Surface par, $r=R$